欧拉回路输出（DFS，不用回溯！）Watchcow POJ 2230

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 8109		Accepted: 3551		Special Judge

Description

Bessie's been appointed the new watch-cow for the farm. Every night, it's her job to walk across the farm and make sure that no evildoers are doing any evil. She begins at the barn, makes her patrol, and then returns to the barn when she's done.

If she were a more observant cow, she might be able to just walk each of M (1 <= M <= 50,000) bidirectional trails numbered 1..M between N (2 <= N <= 10,000) fields numbered 1..N on the farm once and be confident that she's seen everything she needs to see. But since she isn't, she wants to make sure she walks down each trail exactly twice. It's also important that her two trips along each trail be in opposite directions, so that she doesn't miss the same thing twice.

A pair of fields might be connected by more than one trail. Find a path that Bessie can follow which will meet her requirements. Such a path is guaranteed to exist.

Input

* Line 1: Two integers, N and M.

* Lines 2..M+1: Two integers denoting a pair of fields connected by a path.

Output

* Lines 1..2M+1: A list of fields she passes through, one per line, beginning and ending with the barn at field 1. If more than one solution is possible, output any solution.

Sample Input

4 5
1 2
1 4
2 3
2 4
3 4

Sample Output

1
2
3
4
2
1
4
3
2
4
1

Hint

OUTPUT DETAILS:

Bessie starts at 1 (barn), goes to 2, then 3, etc...

Source

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<sstream>
#include<algorithm>
#include<queue>
#include<deque>
#include<iomanip>
#include<vector>
#include<cmath>
#include<map>
#include<stack>
#include<set>
#include<fstream>
#include<memory>
#include<list>
#include<string>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
#define MAXN 10009
#define N 50009
#define MOD 10000007
#define INF 1000000009
const double eps = 1e-;
const double PI = acos(-1.0);

struct edge
{
    edge(int _v, bool _vis) :v(_v), vis(_vis){}
    int v;
    bool vis;
};
vector<edge> E[MAXN];
int n, m;
void DFS(int cur)
{
    for (int i = ; i < E[cur].size(); i++)
    {
        if (!E[cur][i].vis)
        {
            E[cur][i].vis = true;
            DFS(E[cur][i].v);
        }
    }
    printf("%d\n", cur);
}
int main()
{
    while (scanf("%d%d", &n, &m) != EOF)
    {
        for (int i = ; i <= n; i++)
            E[i].clear();
        int f, t;
        for (int i = ; i < m; i++)
            scanf("%d%d", &f, &t), E[f].push_back(edge(t,false)), E[t].push_back(edge(f,false));
        DFS();
    }
}

上面是有向图的回溯

下面是无向图。从度最大的点往前回溯

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<vector>
#include<string>
#include<map>
#include<cstring>
using namespace std;
#define INF 0x3f3f3f3f
#define MAXN 55
#define N 2000
typedef long long LL;

/*
无向图的欧拉回路
从度最大的点开始回溯
*/
int T, n;
int g[MAXN][MAXN];
int degree[MAXN];
void dfs(int k)
{
    for (int i = ; i <= ; i++)
    {
        if (g[k][i])
        {
            g[k][i]--, g[i][k]--;
            dfs(i);
            printf("%d %d\n", i, k);
        }
    }
}
int main()
{
    scanf("%d", &T);
    for (int cas = ; cas <= T; cas++)
    {
        memset(degree, , sizeof(degree));
        memset(g, , sizeof(g));
        scanf("%d", &n);
        for (int i = ; i < n; i++)
        {
            int a, b;
            scanf("%d%d", &a, &b);
            g[a][b]++, g[b][a]++;
            degree[a]++, degree[b]++;
        }
        int Max = -, k = -;
        bool f = true;
        for (int i = ; i < MAXN; i++)
        {
            if (degree[i] > Max)
            {
                Max = degree[i], k = i;
            }
            if (degree[i] %  == )
            {
                f = false;
                break;
            }
        }
        printf("Case #%d\n", cas);
        if (f)
            dfs(k);
        else
            printf("some beads may be lost\n");
        if (cas <= T)
            printf("\n");

    }
}